If two Hermitian matrices anticommute, $MN=-NM$, then if we diagonalize $M$, does $N$ take a particular form? Block antidiagonal, or strictly antidiagonal? From a few really old answers on the site this is what I have gathered: When $M^2=N^2=\mathbb{I}$ (Identity), then M can be written as $$ M = \left(\begin{array}{cc} \mathbb{I} & 0 \\ 0 & -\mathbb{I} \end{array}\right) $$ and $N$ can be brought to a form $$ N = \left(\begin{array}{cc} 0 & B \\ B^{-1} & 0 \end{array}\right) $$ In this answer, the block $B$ is supposed to be 'similar to' Identity on the antidiagonal blocks, but I fail to see how and no explanation is provided, although asked for in the comments.
What happens when we do not have the assumption $N^2=\mathbb{I}$? Is there a way to bring the matrix $N$ to a strictly antidiagonal form?
EDIT: This is for a research project, and I am unable to find this in literature. We will of course cite the solution given here.